The daily exchange rates for the five-year period 2003 to 2008 between the euro (EUR) and the British pound (GBP) are well-modeled by a normal distribution with mean 1.459 euros (to pounds), and standard deviation 0.033 euros. What is the probability that on a randomly selected day during this period, the pound was worth (1) less than 1.459 euros, (2) more than 1.492 euros
When trying to solve this, I did:
1) z=(1.495-1.495)/0.033 and then matched the answer of 0 with z<0 on the standard normal probabilities table to get 0.5, which is correct.
2)(1.492-1.495)/0.033 and got the answer 0.0909, with which I matched it with 0.09 on the table and got 0.4641, but the correct answer is 0.1587 and I am confused about how to get this answer.
Any help is much appreciated!
Let the random variable $X=$The price of the pound in euros on a randomly selected day within this period.
Then
$$X\sim N\left(1.459,0.033^2\right)$$ $$P(X \ge 1.492)=P\left(\cfrac{1.459 - 99}{0.033} \le X \le \cfrac{1.459-1.492}{0.033}\right)=P(-2955 \le z \le -1)=\color{red}{\fbox{0.158655253}}$$ as required.
If you are wondering where the $99$ came from I simply selected an arbitrarily large value to ensure all values of $X$ were covered up to $X=1.492$.
As you are using tables you can simply select a very large and negative $z$ value.